Jordan Block of a complex matrix, with $A^4=I$

Question

The following statement is false or true:

If $A \in M(n, \mathbb{C})$ is a matrix with complex entries of order $n$ such that $A^4=I$ then \begin{pmatrix} i & 1\\ 0 & i \end{pmatrix} Can be a Jordan Block of $A$.

I believe this statement is false, but I could not formalize the demonstration.

Answer 1

Observe that

$$A^4=I\implies (A-I)(A+I)(A^2+I)=(A-I)(A+I)(A-iI)(A+iI)$$

Thus, over $\;\Bbb C\;$ , the matrix's minimal polynomial decomposes as a product of different linear factors and is thus diagonalizable, which means it cannot have a Jordan Block as the one you wrote.

Answer 2

An alternative approach: we have $B = iI + N$, where $I$ denotes the identity matrix and we have $$ N = \pmatrix{0&1\\0&0}. $$ Notably, $iI$ and $N$ commute (so the binomial theorem applies) and $N^2 = 0$. Thus, we have $$ B^4 = (iI + N)^4 = (iI)^4 + 4(iI)^3N + 6(0) + 4(0) + 0 = I -4i N\\ = \pmatrix{1&-4i\\0&1}. $$ Thus, if $B$ is a Jordan block of the Jordan form $J$ of $A$, then $J^4$ will have $B^4$ as a diagonal submatrix. So, we cannot have $J^4 = I$. It follows that $A^4 \neq I$.